Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\mu$ be any probability measure on the interval $]0,\infty[$. I think the following limit holds, but I don't manage to prove it: $$\frac{1}{\alpha}\log\biggl(\int_0^\infty\! x^\alpha d\mu(x)\biggr) \ \xrightarrow[\alpha\to 0+]{}\ \int_0^\infty\! \log x\ d\mu(x)$$ In probabilistic terms it can be rewritten as: $$\frac{1}{\alpha}\log\mathbb{E}[X^\alpha] \ \xrightarrow[\alpha\to 0+]{}\ \mathbb{E}[\log X]$$ for any positive random variable $X$.

Can you help me to prove it?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

We assume that there is $\alpha_0>0$ such that $\int_0^{+\infty}x^{\alpha_0} d\mu(x)$ is finite. Let $I(\alpha):=\frac 1{\alpha}\log\left(\int_0^{+\infty}x^\alpha d\mu(x)\right)$ and $I:=\int_0^{+\infty}\log xd\mu(x)$.

Since the function $t\mapsto \log t$ is concave, we have $I(\alpha)\geqslant I$ for all $\alpha$.

Now, use the inequality $\log(1+t)\leqslant t$ and the dominated convergence theorem to show that $\lim_{\alpha\to 0^+}\int_0^{+\infty}\frac{x^\alpha-1}\alpha d\mu(x)=I$. Call $J(\alpha):=\int_0^{+\infty}\frac{x^\alpha-1}\alpha d\mu(x)$. Then $$I\leqslant I(\alpha)\leqslant J(\alpha).$$

share|improve this answer
    
I agree with your statements. But then how can I use then to prove that $I(\alpha)\xrightarrow[\alpha\to 0+]{}I$? –  qwertyuio Feb 27 '13 at 11:13
    
I've added details. –  Davide Giraudo Feb 27 '13 at 11:24
1  
And the inequality $I(\alpha)\leq J(\alpha)$ holds since in general $\log y\leq y-1$. Ok, now it's clear: thank you very much! –  qwertyuio Feb 27 '13 at 11:44

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.